Prove that the locus of the poles of chords which are normal to the parabola `y^(2) = 4ax` is the curve

Answer

To prove that the locus of the poles of chords which are normal to the parabola \(y^2 = 4ax\) is a specific curve, we will follow these steps:

### Step 1: Understand the parabola and the concept of the pole
The given parabola is \(y^2 = 4ax\). The pole of a chord is a point from which the chord can be drawn, and the chord is normal to the parabola at some point.

### Step 2: Define the pole and the normal chord
Let the pole be at the point \((h, k)\). The equation of the polar line corresponding to this pole with respect to the parabola can be expressed as:
\[ yk = 2ax + \frac{h}{2} \]
This is derived from the general equation of the polar line for a conic.

### Step 3: Simplify the polar equation
Rearranging the polar equation gives:
\[ yk = 2ax + h \]
This can be rewritten as:
\[ yk - 2ax - h = 0 \]

### Step 4: Write the equation of the normal to the parabola
The equation of the normal to the parabola \(y^2 = 4ax\) at a point \((at^2, 2at)\) is given by:
\[ y = mx - 2am + at^3 \]
where \(m\) is the slope of the normal.

### Step 5: Set the two equations equal
Since the polar line and the normal line represent the same line, we can equate their coefficients. Thus, we have:
\[ yk = 2ax + h \quad \text{and} \quad y = mx - 2am + at^3 \]

### Step 6: Compare coefficients
From the two equations, we can compare the coefficients of \(x\) and \(y\):
1. Coefficient of \(x\): \(2a\) (from polar) and \(m\) (from normal)
2. Coefficient of \(y\): \(k\) (from polar) and \(1\) (from normal)

This gives us:
\[ \frac{1}{k} = \frac{m}{2a} \]
From this, we can express \(m\) in terms of \(k\):
\[ m = \frac{2a}{k} \]

### Step 7: Substitute \(m\) back into the equation for \(h\)
From the earlier comparison, we also have:
\[ h = -2a + am^2 \]
Substituting \(m = \frac{2a}{k}\) into this equation gives:
\[ h = -2a + a\left(\frac{2a}{k}\right)^2 \]
Simplifying this results in:
\[ h = -2a + \frac{4a^3}{k^2} \]

### Step 8: Rearranging the equation
Rearranging the equation leads to:
\[ h + 2a = \frac{4a^3}{k^2} \]
Multiplying both sides by \(k^2\) gives:
\[ k^2(h + 2a) = 4a^3 \]

### Step 9: Final form of the locus
This can be rearranged to form the equation of the locus:
\[ hk^2 + 2ak^2 - 4a^3 = 0 \]
This represents the locus of the poles of the normal chords to the parabola.

### Conclusion
Thus, we have shown that the locus of the poles of chords which are normal to the parabola \(y^2 = 4ax\) is indeed a specific curve, as derived above.

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